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Proof step 13 from proof index-update-adr-in (previous step)

# kx0 = # ax1, adr ∈ ri[key, adr1][kx0, ax1], adr ∈ ri[key, adr1] \ kx0.toSet ⊦ adr = adr1, dups(kx0), adr ∈ (ri \ kx0.toSet) -- key, adr ∈ ax1, key ∈ kx0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ adr ∈ ri ↔ (∃ key. key ∈ ri ∧ ri[key] = adr)
Proof step 15:
# kx0 = # ax1, adr ∈ ri[key, adr1] \ kx0.toSet, adr ∈ ri[key, adr1][kx0, ax1] ⊦ adr = adr1, dups(kx0), key ∈ kx0, adr ∈ ax1, ∃ key0. key0 ∈ (ri \ kx0.toSet) -- key ∧ ((ri \ kx0.toSet) -- key)[key0] = adr

Rule argument

spec: store+set, inst: node-index name: Data-in seq: ⊦ adr ∈ ri ↔ (∃ key. key ∈ ri ∧ ri[key] = adr) subst: [ri, adr] → [(ri \ kx0.toSet) -- key, adr] flags: <all paths>

Additional information

Used terms: (ri \ kx0.toSet) -- key, adr